Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.4\) .
a.Find the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.[4]
b.Given that \({\text{P}}(A \cup B) = 0.6\) , find \({\text{P}}(A|B)\) .[3]
▶️Answer/Explanation
Markscheme
(i) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1
(ii) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) (M1)
\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\( = 0.3 + 0.4 – 0.12 = 0.58\) A1
[4 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cup B)\)
\( = 0.3 + 0.4 – 0.6 = 0.1\) A1
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\) (M1)
\( = \frac{{0.1}}{{0.4}} = 0.25\) A1
[3 marks]
Examiners report
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.3\) and \({\text{P}}(B) = 0.4\) .
a.Find the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.[4]
b.Given that \({\text{P}}(A \cup B) = 0.6\) , find \({\text{P}}(A|B)\) .[3]
▶️Answer/Explanation
Markscheme
(i) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) = 0.7\) A1
(ii) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) (M1)
\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\( = 0.3 + 0.4 – 0.12 = 0.58\) A1
[4 marks]
\({\text{P}}(A \cap B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cup B)\)
\( = 0.3 + 0.4 – 0.6 = 0.1\) A1
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\) (M1)
\( = \frac{{0.1}}{{0.4}} = 0.25\) A1
[3 marks]
Examiners report
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Most candidates attempted this question and answered it well. A few misconceptions were identified (eg \({\text{P}}(A \cup B) = {\text{P}}(A){\text{P}}(B)\) ). Many candidates were unsure about the meaning of independent events.
Question
Two events A and B are such that \({\text{P}}(A \cup B) = 0.7\) and \({\text{P}}(A|B’) = 0.6\).
Find \({\text{P}}(B)\).
▶️Answer/Explanation
Markscheme
Note: Be aware that an unjustified assumption of independence will also lead to P(B) = 0.25, but is an invalid method.
METHOD 1
\({\text{P}}(A’|B’) = 1 – {\text{P}}(A|B’) = 1 – 0.6 = 0.4\) M1A1
\({\text{P}}(A’|B’) = \frac{{{\text{P}}(A’ \cap B’)}}{{{\text{P}}(B’)}}\)
\({\text{P}}(A’ \cap B’) = {\text{P}}\left( {(A \cup B)’} \right) = 1 – 0.7 = 0.3\) A1
\(0.4 = \frac{{0.3}}{{{\text{P}}(B’)}} \Rightarrow {\text{P(}}B’) = 0.75\) (M1)A1
\({\text{P}}(B) = 0.25\) A1
(this method can be illustrated using a tree diagram)
[6 marks]
METHOD 2
\({\text{P}}\left( {(A \cup B)’} \right) = 1 – 0.7 = 0.3\) A1
\({\text{P}}(A|B’) = \frac{x}{{x + 0.3}} = 0.6\) M1A1
\(x = 0.6x + 0.18\)
\(0.4x = 0.18\)
\(x = 0.45\) A1
\({\text{P}}(A \cup B) = x + y + z\)
\({\text{P}}(B) = y + z = 0.7 – 0.45\) (M1)
\( = 0.25\) A1
[6 marks]
METHOD 3
\(\frac{{{\text{P}}(A \cap B’)}}{{{\text{P}}(B’)}} = 0.6{\text{ (or P}}(A \cap B’) = 0.6{\text{P}}(B’)\) M1
\({\text{P}}(A \cap B’) = {\text{P}}(A \cup B) – {\text{P}}(B)\) M1A1
\({\text{P}}(B’) = 1 – {\text{P}}(B)\)
\(0.7 – {\text{P}}(B) = 0.6 – 0.6{\text{P}}(B)\) M1(A1)
\(0.1 = 0.4{\text{P}}(B)\)
\({\text{P}}(B) = \frac{1}{4}\) A1
[6 marks]
Examiners report
There is a great variety of ways to approach this question and there were plenty of very good solutions produced, all of which required an insight into the structure of conditional probability. A few candidates unfortunately assumed independence and so did not score well.
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = \frac{2}{5},{\text{ P}}(B) = \frac{{11}}{{20}}\) and \({\text{P}}(A|B) = \frac{2}{{11}}\).
(a) Find \({\text{P}}(A \cap B)\).
(b) Find \({\text{P}}(A \cup B)\).
(c) State with a reason whether or not events \(A\) and \(B\) are independent.
▶️Answer/Explanation
Markscheme
(a) \({\text{P}}(A \cap B) = {\text{P}}(A|B) \times P(B)\)
\({\text{P}}(A \cap B) = \frac{2}{{11}} \times \frac{{11}}{{20}}\) (M1)
\( = \frac{1}{{10}}\) A1
[2 marks]
(b) \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\)
\({\text{P}}(A \cup B) = \frac{2}{5} + \frac{{11}}{{20}} – \frac{1}{{10}}\) (M1)
\( = \frac{{17}}{{20}}\) A1
[2 marks]
(c) No – events A and B are not independent A1
EITHER
\({\text{P}}(A|B) \ne {\text{P}}(A)\) R1
\(\left( {\frac{2}{{11}} \ne \frac{2}{5}} \right)\)
OR
\({\text{P}}(A) \times {\text{P}}(B) \ne {\text{P}}(A \cap B)\)
\(\frac{2}{5} \times \frac{{11}}{{20}} = \frac{{11}}{{50}} \ne \frac{1}{{10}}\) R1
Note: The numbers are required to gain R1 in the ‘OR’ method only.
Note: Do not award A1R0 in either method.
[2 marks]
Total [6 marks]
Examiners report
Question
Events \(A\) and \(B\) are such that \({\text{P}}(A) = 0.2\) and \({\text{P}}(B) = 0.5\).
a.Determine the value of \({\text{P}}(A \cup B)\) when
(i) \(A\) and \(B\) are mutually exclusive;
(ii) \(A\) and \(B\) are independent.[4]
b.Determine the range of possible values of \({\text{P}}\left( {A|B} \right)\).[3]
▶️Answer/Explanation
Markscheme
(i) use of \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B)\) (M1)
\({\text{P}}(A \cup B) = 0.2 + 0.5\)
\( = 0.7\) A1
(ii) use of \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\({\text{P}}(A \cup B) = 0.2 + 0.5 – 0.1\)
\( = 0.6\) A1
[4 marks]
\({\text{P}}\left( {A|B} \right) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}}\)
\({\text{P}}\left( {A|B} \right)\) is a maximum when \({\text{P}}(A \cap B) = {\text{P}}(A)\)
\({\text{P}}\left( {A|B} \right)\) is a minimum when \({\text{P}}(A \cap B) = 0\)
\(0 \le {\text{P}}\left( {A|B} \right) \le 0.4\) A1A1A1
Note: A1 for each endpoint and A1 for the correct inequalities.
[3 marks]
Total [7 marks]
Examiners report
This part was generally well done.
Disappointingly, many candidates did not seem to understand the meaning of the word ‘range’ in this context.
Question
\(A\) and \(B\) are two events such that \({\text{P}}(A) = 0.25,{\text{ P}}(B) = 0.6\) and \({\text{P}}(A \cup B) = 0.7\).
a.Find \({\text{P}}(A \cap B)\).[2]
b.Determine whether events \(A\) and \(B\) are independent. [2]
▶️Answer/Explanation
Markscheme
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\)
\({\text{P}}(A \cap B) = 0.25 + 0.6 = 0.7\) M1
\( = 0.15\) A1
[2 marks]
EITHER
\({\text{P}}(A){\text{P}}(B)( = 0.25 \times 0.6) = 0.15\) A1
\( = {\text{P}}(A \cap B)\) so independent R1
OR
\({\text{P}}(A|B) = \frac{{{\text{P}}(A \cap B)}}{{{\text{P}}(B)}} = \frac{{0.15}}{{0.6}} = 0.25\) A1
\( = {\text{P}}(A)\) so independent R1
Note: Allow follow through for incorrect answer to (a) that will result in events being dependent in (b).
[2 marks]
Total [4 marks]
Examiners report
[N/A]
[N/A]
Question
Two events \(A\) and \(B\) are such that \({\text{P}}(A \cap B’) = 0.2\) and \({\text{P}}(A \cup B) = 0.9\).
a.On the Venn diagram shade the region \(A’ \cap B’\).
[1]
b.Find \({\text{P}}(A’|B’)\).[4]
▶️Answer/Explanation
Markscheme
A1
[1 mark]
\(P(A’|B’) = \frac{{P(A’ \cap B’)}}{{P(B’)}}\) (M1)
\(P(B’) = 0.1 + 0.2 = 0.3\) (A1)
\(P(A’ \cap B’) = 0.1\) (A1)
\(P(A’|B’) = \frac{{0.1}}{{0.3}} = \frac{1}{3}\) A1
[4 marks]
Examiners report
Part (a) was well done.
In part (b) some candidates were unable to write down the conditional probability formula. Some then failed to realise that part (a) was designed to help them work out \(P(A’ \cap B’)\) and instead incorrectly assumed independence.
Question
\(A\) and \(B\) are independent events such that \({\text{P}}(A) = {\text{P}}(B) = p,{\text{ }}p \ne 0\).
a.Show that \({\text{P}}(A \cup B) = 2p – {p^2}\).[2]
b.Find \({\text{P}}(A|A \cup B)\) in simplest form.[4]
▶️Answer/Explanation
Markscheme
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\)
\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A){\text{P}}(B)\) (M1)
\( = p + p – {p^2}\) A1
\( = 2p – {p^2}\) AG
[2 marks]
\({\text{P}}(A|A \cup B) = \frac{{{\text{P}}\left( {A \cap (A \cup B)} \right)}}{{{\text{P}}(A \cup B)}}\) (M1)
Note: Allow \({\text{P}}(A \cap A \cup B)\) if seen on the numerator.
\( = \frac{{{\text{P}}(A)}}{{{\text{P}}(A \cup B)}}\) (A1)
\( = \frac{p}{{2p – {p^2}}}\) A1
\( = \frac{1}{{2 – p}}\) A1
[4 marks]
Examiners report
Part (a) posed few problems. Part (b) was possibly a good discriminator for the 4/5 candidates. Some were aware of an alternative (useful) form for the conditional probability, but were unable to interpret \(P\left( {A \cap (A \cup B)} \right)\). Large numbers of fully correct answers were seen.
Part (a) posed few problems. Part (b) was possibly a good discriminator for the 4/5 candidates. Some were aware of an alternative (useful) form for the conditional probability, but were unable to interpret \(P\left( {A \cap (A \cup B)} \right)\). Large numbers of fully correct answers were seen.
Question
Consider two events \(A\) and \(A\) defined in the same sample space.
Given that \({\text{P}}(A \cup B) = \frac{4}{9},{\text{ P}}(B|A) = \frac{1}{3}\) and \({\text{P}}(B|A’) = \frac{1}{6}\),
a.Show that \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A’ \cap B)\).[3]
b.(i) show that \({\text{P}}(A) = \frac{1}{3}\);
(ii) hence find \({\text{P}}(B)\).[6]
▶️Answer/Explanation
Markscheme
METHOD 1
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) M1
\( = {\text{P}}(A) + {\text{P}}(A \cap B) + {\text{P}}(A’ \cap B) – {\text{P}}(A \cap B)\) M1A1
\( = {\text{P}}(A) + {\text{P}}(A’ \cap B)\) AG
METHOD 2
\({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A \cap B)\) M1
\( = {\text{P}}(A) + {\text{P}}(B) – {\text{P}}(A|B) \times {\text{P}}(B)\) M1
\( = {\text{P}}(A) + \left( {1 – {\text{P}}(A|B)} \right) \times {\text{P}}(B)\)
\( = {\text{P}}(A) + {\text{P}}(A’|B) \times {\text{P}}(B)\) A1
\( = {\text{P}}(A) + {\text{P}}(A’ \cap B)\) AG
[3 marks]
(i) use \({\text{P}}(A \cup B) = {\text{P}}(A) + {\text{P}}(A’ \cap B)\) and \({\text{P}}(A’ \cap B) = {\text{P}}(B|A’){\text{P}}(A’)\) (M1)
\(\frac{4}{9} = {\text{P}}(A) + \frac{1}{6}\left( {1 – {\text{P}}(A)} \right)\) A1
\(8 = 18{\text{P}}(A) + 3\left( {1 – {\text{P}}(A)} \right)\) M1
\({\text{P}}(A) = \frac{1}{3}\) AG
(ii) METHOD 1
\({\text{P}}(B) = {\text{P}}(A \cap B) + {\text{P}}(A’ \cap B)\) M1
\( = {\text{P}}(B|A){\text{P}}(A) + {\text{P}}(B|A’){\text{P}}(A’)\) M1
\( = \frac{1}{3} \times \frac{1}{3} + \frac{1}{6} \times \frac{2}{3} = \frac{2}{9}\) A1
METHOD 2
\({\text{P}}(A \cap B) = {\text{P}}(B|A){\text{P}}(A) \Rightarrow {\text{P}}(A \cap B) = \frac{1}{3} \times \frac{1}{3} = \frac{1}{9}\) M1
\({\text{P}}(B) = {\text{P}}(A \cup B) + {\text{P}}(A \cap B) – {\text{P}}(A)\) M1
\({\text{P}}(B) = \frac{4}{9} + \frac{1}{9} – \frac{1}{3} = \frac{2}{9}\) A1
[6 marks]
Examiners report
[N/A]
[N/A]